Question

If LCM of the two polynomials (x^2 + 3x)
(x^2 + 3x + 2) and (x^2 + 6x + 8)(x^2 + kx + 6) is
x(x + 1)(x + 2)^2 (x + 3)(x + 4), then find k.​
"please answer fast"​

Answer 1

Answer:

5

first polynomial can be written as (x)(x+3)(x+2)(x+1)

similarly for second polynomial (x+2)(x+4)(x^2 +kx+6)

since LCM is the least common multiple both the polynomials should be able to divide the LCM

so let

(x^2 +kx+6) be factorized as (x+a)(x+b)

now we know

ab =6

a+b=k

(x+a) and (x+b) should be from the factors in the LCM

and to follow the condition ab=6

we could observe that (x+2) and (x+3) are the required ones

ab=2×3

a+b=5

Answer 2

$\mathfrak{\huge \underline{Question:}}$

$If~LCM~of~the~two~polynomials~\\\\ (x^2 + 3x)(x^2 + 3x + 2) ~and~ (x^2 + 6x + 8)(x^2 + kx + 6)\\\\is~x(x + 1)(x + 2)^2 (x + 3)(x + 4), then find ~\bf k.$

$\mathfrak{\huge \underline{Answer:}}$

$\huge \boxed{\fcolorbox{black} {pink}{K=5}}$

$\mathfrak{\huge \underline{Explanation:}}$

Factorize both the polynomial:

$(x^2 + 3x)(x^2 + 3x + 2) = \bf x(x+3)(x+1)(x+2) \\\\(x^2 + 6x + 8)(x^2 + kx + 6)=\bf(x+2)(x+4)(x^2+kx+6) \\\\ LCM~of~both~the~polynomial = x(x+3)(x+1)(x+2)(x+4)(x^2+kx+6).$

But, LCM given is

$x(x + 1)(x + 2)^2 (x + 3)(x+4)$

So,

$x(x+3)(x+1)(x+2)(x+4)(x^2+kx+6)=x(x + 1)(x + 2)^2 (x + 3)(x + 4)\\\\ \cancel{x(x+3)(x+1)(x+2)(x+4)}(x^2+kx+6) =\cancel{x(x + 1)(x + 2)(x + 3)(x + 4)}(x+2) \\\\ x+2 is~ a ~factor ~of ~x^2+kx+6$

So, x+2= 0, x= - 2(by remainder theorem)

Putting the value of $x=-2~ in~ x^2+kx+6,$ we get

4-2k+6=0

10-2k=0

2k=10

k=5